A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

My final answers:
1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.

A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

My final answers:
1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.

A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

These look OK to me. Notationwise, you can write ##\sqrt[4]{-46}## as \sqrt[4]{-46}. The fourth roots of -46 will be arranged equidistant around a circle: at ##\pi/4, 3\pi/4, 5\pi/4, 7\pi/4##.

RJLiberator said:

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

Right. ##x^4 + 46## can't be factored over the reals.

RJLiberator said:

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

My final answers:
1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.

A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.

3. The attempt at a solution

Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

We have x^4+46.
To start, let's just factor it as I normally would without worry about the fields.

With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
(x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

This should be the correct answer to c as it is in completely factored form.

Now, we try to get the factorization over the reals, ℝ. part b:
Since we cant use i, we try to multiply the factors a bit to get rid of i.
We get : (x^2-4throot(-46)) (x^2+4throot(-46))

Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
Answer: Likely not as this is an expression of a+ib.
So, this doesn't work.

At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
But I see to have come to the conclusion that over the field ℝ it is also completely factored.

My final answers:
1.a It is already completely factored
1.b It is completely factored
1.c I factored it above.

As for part 2, the difficulty here is dealing with the mod 5.
If we look at it like 4x^3+2x^2+x+3
I would think it would be in the form:
(x+)(x+)(x+)

But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.[/QUOTE]

Staff: Mentor

So my thoughts were confirmed. Part 1 cannot be factored over Q or R. Is there any way I should show this?

Sure. Suppose that ##x^4 + 46 = 0##. Clearly x = 0 is not a solution. If x > 0, there are clearly no solutions, as both ##x^4## and 46 are positive. The same is true if x < 0. This means there are no solutions to the equation, and hence, no real factors of ##x^4 + 46##.

If there are no real factors, then a fortiori, there can't be rational factors, either.

No.
You have a polynomial in ##\mathbb Z_5[x]##.
You want to find the zeros in ##\mathbb Z_5## of that polynomial.
There may be fancy methods to do that, but as ##\mathbb Z_5## contains 5 elements, you could just evaluate the polynomial in each of these 5 elements to find all the zeros.

Ah, I see what you are saying, so
let x be 1, 2, 3, 4, 5 and see which one makes the equation true:
4x^3+2x^2+x+3 = 0

So we have x = 1, 2, and 4 that make this equation true.

But, er, not sure where to go from here.

We found the roots.

Now do what one always does: if ##a## is a root of a polynomial ##p## of degree 3, then ##p(x)=(x-a)q(x)##, where ##q## is a polynomial of degree 2.
And so on.
Just remember that you are working in ##\mathbb Z_5##.
Since you found 3 roots (I found the same 3 roots, so that looks good), you are almost done.

@Samy_A
I finally understand what you meant, it is obvious and important. Sorry I was so dazed this morning.

Here is what I did.
1. Found the roots of the initial polynomial in mod 5. These were 1, 2, and 4.
2. I divided the initial polynomial by one of the roots, I choose (x-1). I was sure to use the mod 5 distinction here.
3. I found that (x-1)(4x^2+x+2) was a factor.
4. I repeated this process for the second term.

I ended up with the answer 4(x+1)(x-1)(x-2) as the completely factored form.